# $f^{-1}(H)$ and $H$ have same codimension

Let $E$ and $F$ be two real finite-dimensional vector spaces. Let $f \, : \, E \, \rightarrow \, F$ be a linear transformation from $E$ to $F$ and $H$ a vector subspace of $F$ such that $F = H + \mathrm{Im}(f)$. I would like to prove that $f^{-1}(H)$ and $H$ have the same codimension.

By definition, $\mathrm{codim}(H) = \dim(F) - \dim(H)$. Showing that $f^{-1}(H)$ has the same codimension as $H$ is equivalent to finding/constructing a subspace $G$ of $E$ such that $E = G \oplus f^{-1}(H)$ and $\dim(G) = \dim(F)-\dim(H)$. How can I construct such a subspace $G$ ?

Another idea: the Grassman formula gives : $$\dim(F) = \dim(H) + \mathrm{rg}(f) - \dim \big( H \cap \mathrm{Im}(f) \big).$$ So, the codimension of $H$ is also equal to $\mathrm{rg}(f) - \dim\big( H \cap \mathrm{Im}(f) \big)$. So $\dim(G)$ should satisfy to: $\dim(G) + \dim\big( H \cap \mathrm{Im}(f) \big) = \mathrm{rg}(f)$.

1. The transformation $f: E \to F$ induces an injective linear transformation $$\widetilde{f}: E/f^{-1}(H) \to F/H$$
2. The map $\widetilde{f}$ is surjective (here you need your assumption $F = H + Im(f)$).
3. Parts 1. and 2. imply that $\widetilde{f}$ is an isomorphism, so the dimensions of domain and target are equal. Since $\dim(E/f^{-1}(H)) = \text{codim}_E(f^{-1}(H))$ and $\dim(F/H) = \text{codim}_F(H)$ you are done.
• Thank you for your answer. I do not get why $\tilde{f}$ is injective. Oct 2, 2017 at 13:41
• In my opinion, it is equivalent (up to isomorphism) to consider $\tilde{f} \, : \, G \, \rightarrow \, K$ where $G,K$ are subspaces such that $E =G \oplus f^{-1}(H)$ and $F = K \oplus H$. But in that case, unless I am mistaken, $\mathrm{ker}\big( \tilde{f} \big) = \mathrm{ker}(f) \cap G$ and I do not see why $\mathrm{ker}(f) \cap G = \lbrace 0 \rbrace$. Oct 2, 2017 at 13:50
• You have $\widetilde{f}(x + f^{-1}(H)) = f(x) + H$ by definition and this is zero in $F/H$ if and only if $f(x) \in H$, that is $x \in f^{-1}(H)$ and so $x + f^{-1}(H)$ was zero in $E/f^{-1}(H)$ to begin with. Oct 2, 2017 at 14:22
• Your approach also works, just note that $\ker(f) = f^{-1}(\{0\}) \subseteq f^{-1}(H)$, so $G \cap \ker(f) = \{0\}$ as desired. Oct 2, 2017 at 14:27